Question: $f(x) = \dfrac{ 6 }{ \sqrt{ 9 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
Solution: First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $9 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 9$ , which means $-9 \leq x \leq 9$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 9 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 9$ This means that $x \neq 9$ and $x \neq -9$ So we have four restrictions: $x \geq -9$ $x \leq 9$ $x \neq -9$ , and $x \neq 9$ Combining these four, we know that $x > -9$ and $x < 9$ ; alternatively, that $-9 < x < 9$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -9< x <9\, \}$.